Problem E7 of Richard Guy's "Unsolved problems in number theory"

Batalov · 2013-03-13, 03:43

Quote:

Originally Posted by Tomás Oliveira e Silva (NMBRTHRY@LIST)

Richard K. Guy's "Unsolved problems in number theory" problem E7 asks if $\sum_n (-1)^n n/p_n$ , with $p_n$ the n-th prime number, converges or diverges (a question asked by P. Erdos). In its 2004 third edition, no references related to this problem were given.

Has anyone worked on it after 2003?

Preliminary numerical experiments suggest that the sum converges (slowly, of course) to a value close to -0.05216.

Is anyone interested to think about this problem?

LaurV · 2013-03-13, 04:45

When one approximates n/Pn with 1/log, then it seems that it converges. No idea how to prove, tho... You ask me to prove it if Erdos could not?! :surprised

only_human · 2013-03-13, 07:51

Quote:

Originally Posted by LaurV

You ask me to prove it if Erdos could not?! :surprised

Seems worthwhile to talk about.

Richard K. Guy replied:

Quote:

I haven't received any further information about this, but would be glad to learn if there is any. R.

science_man_88 · 2013-03-13, 12:20

Quote:

Originally Posted by Batalov

Is anyone interested to think about this problem?

I'm not really mathematical enough to think about it to a great degree but:

Quote:

Originally Posted by princeps

https://primenumberformula.wordpress.com/

gives multiple formula for $p_n$

could these lead to a simplification that might have a easy way to solve ?

only_human · 2013-03-13, 18:42

To help frame this question about convergence, I've included the text as stated in Richard Guy's book:

Quote:

If p_n is the n th prime, Erdős asks if ∑ (-1)ⁿ n/p_n converges. He notes that the series ∑ (-1)ⁿ (n ln n)/p_n diverges.

Batalov · 2013-03-13, 19:25

I reposted the message from NMBRTHRY, because it looked like a fantastic exercise for any skill level (except for a very low one. Think of it as a projectEuler-on-steroids problem, -- only the answer is not known). For a technically minded audience (myself likely included; I have no relevant theoretical knowledge), there's an interesting aspect of summation with error accumulation estimate and control. The zero level thinking would be that "surely, the positive and negative errors will cancel on average"; one needs to depart from that at least one level higher, and get a bound for expected error, or else after a while the summation will become meaningless. The sequence grows very, very slowly.

I've tinkered for half an hour to see where I could get naïvely; ran some simple-minded calculation with a 40-liner and I was barely able to see the value -0.0304 surpassed. (So, that's ways and ways from Oliveira e Silva's value! And it is growing so slow, that I cannot yet imagine what modeling he used...) I run summation in pairs of terms, with the parwise sum rewritten to minimize error. I left it for myslef for the weekend to hijack yafu sieve and refactor with the strength of yafu code behind this. Could be fun for anyone, I thought, ...and that's when I reposted.

only_human · 2013-03-13, 19:35

Quote:

Originally Posted by Batalov

I reposted the message from NMBRTHRY, because it looked like a fantastic exercise for any skill level (except for a very low one. (...//...) Could be fun for anyone, I thought, ...and that's when I reposted.

I agree wholeheartedly.

R.D. Silverman · 2013-03-13, 20:08

Quote:

Originally Posted by Batalov

I reposted the message from NMBRTHRY, because it looked like a fantastic exercise for any skill level

Actually, the problem is much subtler than it seems. The difficulty is
showing that the series does not "diverge by oscillation".

This, in turn, involves showing that there do not exist too many intervals
in which the primes are somewhat denser than average, followed by
intervals in which the primes are somewhat less dense than average.

Quote:

The zero level thinking would be that "surely, the positive and negative errors will cancel on average"; one needs to depart from that at least one level higher, and get a bound for expected error, or else after a while the summation will become meaningless. The sequence grows very, very slowly.

Indeed.

The problem is showing that the errors are essentially a 'random walk';
that they oscillate positive and negative, and do not have long runs
(when primes are too close (or sparse) on average) where the error
remains either positive or negative for long periods.

The answer depends on very subtle issues in the distribution of primes.

ishkibibble · 2013-03-13, 21:28

Questions regarding the distribution of primes are always interesting.
(As an analogy, looking at problems like this with a practiced eye helps. The example of Marion Tinsley shows that you should have a feeling about what approach will lead to a desired outcome.)
I do have some interesting results which may apply to this question but at present I have more questions than answers. A recent prime I posted is a test value for a certain line of inquiry that is associated with this limit process.
I'll have more to post if successful.

Batalov · 2013-03-13, 22:38

I have sieved to p(n)<30e9 (that's up to n<1300005928; it was a toy sieve, and without bit packing I blew up memory already; need to refactor with yafu or A.Kulsha's program?; need to walk in windows after that) and the growth is slower than ln ln n. Of course, it can oscillate on much larger scale. Oliveira e Silva suggests that it converges to -0.05216, which is on my scale (so far) nowhere near in sight...

alpertron · 2013-03-14, 13:24

This is my attempt at the conditional proof:

We will perform the sum two terms at a time so we will always add positive numbers.

On the Riemann hypothesis:

$p_{n+1}-p_n = O(\sqrt{p_n}\,\log p_n)$ (see http://en.wikipedia.org/wiki/Cram%C3%A9r%27s_conjecture )

$\frac{1}{p_n} \,- \,\frac{1}{p_n + O(\sqrt{p_n}\,\log p_n)}\, < \,\frac{k(\sqrt{p_n}\,\log p_n)}{p_n^2}\, < \,\frac{k}{p_n^{1.5 - \epsilon}}\, < \,\frac{k}{n^{1.5 - \epsilon}$

$\sum_{i=1}^{\infty} \frac{k}{{(2i)}^{1.5 - \epsilon}$ is convergent so the original sum must be convergent as well.

PD: Forget everything I wrote. I understood that the summand was $(-1)^n/p_n$ , not $(-1)^n\,n/p_n$ as posted by the OP.

2013-03-13, 04:45	#2
LaurV Romulan Interpreter "name field" Jun 2011 Thailand 3·23·149 Posts	When one approximates n/Pn with 1/log, then it seems that it converges. No idea how to prove, tho... You ask me to prove it if Erdos could not?! :surprised Last fiddled with by LaurV on 2013-03-13 at 04:46

2013-03-13, 21:28	#9
ishkibibble Nov 2012 Canada 10101₂ Posts	Pn Questions regarding the distribution of primes are always interesting. (As an analogy, looking at problems like this with a practiced eye helps. The example of Marion Tinsley shows that you should have a feeling about what approach will lead to a desired outcome.) I do have some interesting results which may apply to this question but at present I have more questions than answers. A recent prime I posted is a test value for a certain line of inquiry that is associated with this limit process. I'll have more to post if successful.

2013-03-13, 22:38	#10
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 10,061 Posts	I have sieved to p(n)<30e9 (that's up to n<1300005928; it was a toy sieve, and without bit packing I blew up memory already; need to refactor with yafu or A.Kulsha's program?; need to walk in windows after that) and the growth is slower than ln ln n. Of course, it can oscillate on much larger scale. Oliveira e Silva suggests that it converges to -0.05216, which is on my scale (so far) nowhere near in sight... Attached Thumbnails Last fiddled with by Batalov on 2013-03-13 at 22:45 Reason: sorry, 30e9; with bit packing can sieve easily to 1e12, of course

2013-03-14, 13:24	#11
alpertron Aug 2002 Buenos Aires, Argentina 3·499 Posts	This is my attempt at the conditional proof: We will perform the sum two terms at a time so we will always add positive numbers. On the Riemann hypothesis: $p_{n+1}-p_n = O(\sqrt{p_n}\,\log p_n)$ (see http://en.wikipedia.org/wiki/Cram%C3%A9r%27s_conjecture ) $\frac{1}{p_n} \,- \,\frac{1}{p_n + O(\sqrt{p_n}\,\log p_n)}\, < \,\frac{k(\sqrt{p_n}\,\log p_n)}{p_n^2}\, < \,\frac{k}{p_n^{1.5 - \epsilon}}\, < \,\frac{k}{n^{1.5 - \epsilon}$ $\sum_{i=1}^{\infty} \frac{k}{{(2i)}^{1.5 - \epsilon}$ is convergent so the original sum must be convergent as well. PD: Forget everything I wrote. I understood that the summand was $(-1)^n/p_n$ , not $(-1)^n\,n/p_n$ as posted by the OP. Last fiddled with by alpertron on 2013-03-14 at 13:38

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2013-03-13, 19:25	#6
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 274D₁₆ Posts	I reposted the message from NMBRTHRY, because it looked like a fantastic exercise for any skill level (except for a very low one. Think of it as a projectEuler-on-steroids problem, -- only the answer is not known). For a technically minded audience (myself likely included; I have no relevant theoretical knowledge), there's an interesting aspect of summation with error accumulation estimate and control. The zero level thinking would be that "surely, the positive and negative errors will cancel on average"; one needs to depart from that at least one level higher, and get a bound for expected error, or else after a while the summation will become meaningless. The sequence grows very, very slowly. I've tinkered for half an hour to see where I could get naïvely; ran some simple-minded calculation with a 40-liner and I was barely able to see the value -0.0304 surpassed. (So, that's ways and ways from Oliveira e Silva's value! And it is growing so slow, that I cannot yet imagine what modeling he used...) I run summation in pairs of terms, with the parwise sum rewritten to minimize error. I left it for myslef for the weekend to hijack yafu sieve and refactor with the strength of yafu code behind this. Could be fun for anyone, I thought, ...and that's when I reposted.